Let $g$ be a vector-valued function defined by $g(t)=(\sin(\pi t),-2\cos(\pi t))$. Find $g'(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $(-\sin(\pi t),-2\sin(\pi t))$ (Choice B) B $\cos(\pi t)+2\sin(\pi t))$ (Choice C) C $(-\cos(\pi t),2\sin(\pi t))$ (Choice D) D $(\pi\cos(\pi t),2\pi\sin(\pi t))$
Answer: $g$ is a vector-valued function. This means it takes one number as an input $(t)$, but it outputs two numbers as a two-dimensional vector. Finding the derivative of a vector-valued function is pretty straightforward. Suppose a vector-valued function is defined as $u(t)=(v(t),w(t))$, then its derivative is the vector-valued function $u'(t)=(v'(t),w'(t))$. In other words, the derivative is found by differentiating each of the expressions in the function's output vector. Recall that $g(t)=(\sin(\pi t),-2\cos(\pi t))$. Let's differentiate the first expression: $\dfrac{d}{dt}[\sin(\pi t)]=\pi\cos(\pi t)$ Let's differentiate the second expression: $\dfrac{d}{dt}[-2\cos(\pi t)]=2\pi\sin(\pi t)$ Now let's put everything together: $\begin{aligned} g'(t)&=\left(\dfrac{d}{dt}[\sin(\pi t)],\dfrac{d}{dt}[-2\cos(\pi t)]\right) \\\\ &=(\pi\cos(\pi t),2\pi\sin(\pi t)) \end{aligned}$ In conclusion, $g'(t)=(\pi\cos(\pi t),2\pi\sin(\pi t))$.